Question

Find the principal value of \[{{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)\]

Answer 1

Hint: let us consider the y as given inverse trigonometric function and then we will get the value of \[\cos y\] and if \[\cos y\] is negative then the principal value will be \[\pi -\theta \]. The principal value of \[\cos \theta \] lies between 0 and \[\pi \]

Complete step-by-step answer:
Let y=\[{{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[\Rightarrow \cos y=\dfrac{-\sqrt{3}}{2}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Since \[\dfrac{-\sqrt{3}}{2}\]is negative, principal value of \[\theta \] is \[\pi -\theta \]
\[\Rightarrow \cos y=\dfrac{-\sqrt{3}}{2}=\cos \left( \pi -\dfrac{\pi }{6} \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
We know the principal value of \[{{\cos }^{-1}}\]is \[\left[ 0,\pi \right]\]
Hence the principal value of \[{{\cos }^{-1}}\left( \dfrac{-\sqrt{3}}{2} \right)\]is
\[\pi -\dfrac{\pi }{6}=\dfrac{6\pi -\pi }{6}=\dfrac{5\pi }{6}\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (4)

Note: The general formula for principal value of \[{{\cos }^{-1}}\left( \sin \theta \right)=\dfrac{\pi }{2}-\theta \] if and only if \[\theta \in \left[ -\dfrac{\pi }{2},\dfrac{\pi }{2} \right]\]. The principal value of \[\theta \] for any given inverse function should lie within its range. The range of inverse cos function or arc cos function is \[\left[ 0,\pi \right]\]. So the principle of \[\theta \] for inverse cosine function always lies between 0 and \[\pi \].